Question: Find $\lim_{x\to\infty}\dfrac{\sin(x)}{9-x^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $-3$ (Choice C) C $0$ (Choice D) D The limit doesn't exist
Solution: When dealing with limits that include $\sin(x)$, it's important to remember that $\lim_{x\to\infty}\sin(x)$ doesn't exist, as $\sin(x)$ keeps oscillating between $-1$ and $1$ forever. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ $y$ $x$ $y=\sin(x)$ This doesn't necessarily mean that our limit doesn't exist. Think what happens to $\dfrac{\sin(x)}{9-x^2}$ as $x$ increases towards positive infinity. $\sin(x)$ oscillates between $-1$ and $1$. This can be represented mathematically by the following double inequality: $-\dfrac{1}{9-x^2}\leq\dfrac{\sin(x)}{9-x^2}\leq\dfrac{1}{9-x^2}$ The result is a graph that's always between the graphs of $y=\pm\dfrac{1}{9-x^2}$ (the dashed lines). ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}2}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${0.5}$ ${\llap{-}0.5}$ $y$ $x$ Since $\lim_{x\to\infty}\pm\dfrac{1}{9-x^2}=0$, so must our limit be equal to $0$. (This is similar to the Squeeze Theorem, only applied at infinity.) In conclusion, $\lim_{x\to\infty}\dfrac{\sin(x)}{9-x^2}=0$.